Simplify; express your answer in exponential form. Assume $q\neq 0, x\neq 0$. $\dfrac{{(q^{-3})^{-5}}}{{(q^{5}x^{-2})^{-4}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{-3}}$ to the exponent ${-5}$ . Now ${-3 \times -5 = 15}$ , so ${(q^{-3})^{-5} = q^{15}}$ In the denominator, we can use the distributive property of exponents. ${(q^{5}x^{-2})^{-4} = (q^{5})^{-4}(x^{-2})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{-3})^{-5}}}{{(q^{5}x^{-2})^{-4}}} = \dfrac{{q^{15}}}{{q^{-20}x^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{15}}}{{q^{-20}x^{8}}} = \dfrac{{q^{15}}}{{q^{-20}}} \cdot \dfrac{{1}}{{x^{8}}} = q^{{15} - {(-20)}} \cdot x^{- {8}} = q^{35}x^{-8}$.